(a) Simplify : \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\) (b) Solve the following equations correct to one decimal place. (i) \(\tan (\theta + 25)...
Question 1 Report
(a) Simplify : \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\)
(b) Solve the following equations correct to one decimal place.
(i) \(\tan (\theta + 25)° = 5.145\)
(ii) \(5\cos \theta - 1 = 0\), where \(0° \leq \theta \leq 90°\).
Answer Details
(a) Write \(625=5^{4}\) and \(25=5^{2}\).
\[625^{\frac{3}{8}}\times5^{\frac{1}{2}}\div25=(5^{4})^{\frac{3}{8}}\times5^{\frac{1}{2}}\div5^{2}=5^{\frac{12}{8}}\times5^{\frac{1}{2}}\times5^{-2}\]Add the indices: \(\dfrac{12}{8}+\dfrac{1}{2}-2=\dfrac{3}{2}+\dfrac{1}{2}-2=2-2=0\).
\[=5^{0}=1\]\(=1\)
(b)(i) \(\tan(\theta+25)^{\circ}=5.145\).
\((\theta+25)^{\circ}=\tan^{-1}5.145=79.0^{\circ}\), so \(\theta=79.0-25=54.0^{\circ}\).
\(\theta\approx 54.0^{\circ}\)
(ii) \(5\cos\theta-1=0\Rightarrow \cos\theta=\dfrac{1}{5}=0.2\).
\(\theta=\cos^{-1}0.2=78.5^{\circ}\) (within \(0^{\circ}\le\theta\le90^{\circ}\)).
\(\theta\approx 78.5^{\circ}\)
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